The mid-P binomial test for the binomial probability (pi)

The mid-P binomial test for the binomial probability (pi) H_0: pi = pi0 vs H_A: pi ~= pi0 (two-sided) Described in Chapter 2 "The 1x2 Table and the Binomial Distribution"

MidP_binomial_test_1x2(X, n, pi0)

Arguments

X

the number of successes

n

the total number of observations

pi0

a given probability

Value

An object of the contingencytables_result class, basically a subclass of base::list(). Use the utils::str() function to see the specific elements returned.

Examples

# The number of 1st order male births (Singh et al. 2010, adapted)
MidP_binomial_test_1x2(singh_2010["1st", "X"], singh_2010["1st", "n"], pi0 = .5)
#> The mid-P binomial test: P = 0.15321
# The number of 2nd order male births (Singh et al. 2010, adapted)
MidP_binomial_test_1x2(singh_2010["2nd", "X"], singh_2010["2nd", "n"], pi0 = .5)
#> The mid-P binomial test: P = 0.84399
# The number of 3rd order male births (Singh et al. 2010, adapted)
MidP_binomial_test_1x2(singh_2010["3rd", "X"], singh_2010["3rd", "n"], pi0 = .5)
#> The mid-P binomial test: P = 0.00252
# The number of 4th order male births (Singh et al. 2010, adapted)
MidP_binomial_test_1x2(singh_2010["4th", "X"], singh_2010["4th", "n"], pi0 = .5)
#> The mid-P binomial test: P = 0.00164
# Ligarden et al. (2010, adapted)
MidP_binomial_test_1x2(ligarden_2010["X"], ligarden_2010["n"], pi0 = .5)
#> The mid-P binomial test: P = 0.01273